PUNJAB PUBLIC SERVICE COMMISSION WRITTEN TEST FOR THE POST OF LECTURER IN MATHEMATICS 2015


PUNJAB PUBLIC SERVICE COMMISSION WRITTEN TEST FOR THE POST OF LECTURER IN MATHEMATICS 2015

Q.01 \fn_cm \int_{-4}^{0}\frac{t}{\sqrt{16-t^2}}dt=__________________

A.   0

B.   Divergent

C.   -4

D.   4

Check Answer
C
View Explanation
  \fn_cm \int_{-4}^{0}\frac{t}{\sqrt{16-t^2}}dt=-\frac{1}{2}\int_{-4}^{0}\frac{-2t}{\sqrt{16-t^2}}dt=-\sqrt{16-t^2}|_{-4}^{0}=-4 

Q.02  The period of function \fn_cm Acos\omega t+Bsin\omega t is

A.  \fn_cm \frac{\omega}{2\pi}

B.  \fn_cm 2\pi \omega

C.  \fn_cm \frac{2\omega}{\pi}

D.  \fn_cm \frac{2\pi}{\omega}

Check Answer
D
View Explanation
 

As sint and cost are periodic function with period 2π
So
Sinwt=sin(wt+2π)=sin(t+2π/w). Similarly for cost . 
Hence period of given function if 2π/w
===> D is correct

Q.03 \inline \fn_cm A= (-4x-3y+az)i+(bx+3y+5z)j+(4x+cy+3z)k is irrotational if a,b,c are__________

A.   4,-3,5

B.   4,5,-3

C.   -3,4,5

D.   2,3,5

Check Answer
D
View Explanation
 

Flow is irrational if \fn_cm \nabla \times A=0\\ \\ \begin{bmatrix} i& j & k\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} &\frac{\partial }{\partial z} \\ (-4x-3y+az)& (bx+3y+5z) & (4x+cy+3z) \end{bmatrix}=0\\ \\(c-5)i-(4-a)j+(b+3)k=0i+0j+0k\\ \\ (a,b,c)=(4,-3,5)

Q.04 \fn_cm V= (-4x-6y+3z)i+(-2x+y-5z)j+(5x+6y+az)k  is solenoidal if a = ________

A.   1

B.   2

C.   3

4.   4

Check Answer
C
View Explanation
 

For solenoidal vector  \fn_cm \nabla.V=0
After applying partial derivative we get
-4+1+a=0
So a=3
C is correct

Q.05 \fn_cm \fn_cm \int_{(0,0)}^{(2,1)} (10x^4-2xy^3)dx-3x^2y^2    along the path \fn_cm x^4-6xy^3=4y^2 is

A.   56

B.   60

C.   62

D.  64 

Check Answer
B
View Explanation
 

We know that 

\fn_cm M=(10x^4-2xy^3),N=-3x^2y^2\\ \\ \\ \frac{\partial M}{\partial y}=-6x^2\\\\\\ \frac{\partial N}{\partial x}=-6x^2

As the solution is exact. So we have

\fn_cm \int_{(0,0)}^{(2,1)} (10x^4-2xy^3)dx-independent \; terms \; of \; x\\\\\\ \int_{(0,0)}^{(2,1)} (10x^4-2xy^3)dx-0 \\\\\\ ={2x^5}-x^2y^3|_{(0,0)}^{(2,1)}\\\\\\ ={64}-{4}=60

Q.06 If S is the closed surface and v is the volume enclosed by S then \fn_cm \int \int_s r.nds=
A. v
B.  2v
C.  3v
D.  4v

Check Answer
C
View Explanation
 

By Gauss Divergence Theorem 

\fn_cm \int \int_s F.\widehat{N}ds=\int \int_r \int dN.Fdv

As position vector \fn_cm \overrightarrow{r}= x i+yj+zk

So  \fn_cm \int \int_s r.nds=\int \int_v \int \nabla.\overrightarrow{r}dv\\\\\\ =\int \int_v \int (\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k).(xi+yj+zk)dv\\\\\\ =\int \int_v \int(1+1+1)dv\\\\\\ =3\int \int_v \int dv =3v

Q.07 Centrifugal acceleration is

A.  \fn_cm -\omega \times (\omega \times r)

B.  \fn_cm \omega \times (\omega \times r)
C.  \fn_cm \omega . (\omega \times r)
D.  \fn_cm r \times (\omega \times r)

Check Answer
A
View Explanation
 

As \fn_cm \overrightarrow{r}= cos \omega t + sin\omega t\\\\\\ \frac {d \overrightarrow{r}}{dt}=\overrightarrow{v}=- \omega sin \omega t +\omega cos\omega t\\\\\\ \frac {d \overrightarrow{v}}{dt}=\overrightarrow{a}=- \omega^2 cos \omega t -\omega^2 sin\omega t\\\\\\ \overrightarrow{a}=-\omega^2 \overrightarrow{r}=-\omega \times( \omega \times \overrightarrow{r} )

Q.08  Number of degrees of freedom of two particles connected by a rigid rod moving treely in a plane Is

A. 2
B.  3
C.  4
D.  5

Check Answer
C
View Explanation
 

No. of degree of freedom= No. of particles × 2

                                              = 2  ×  2  =   4

Q.09  The centrold of a uniform semicircular wire of radius a is

A. \fn_cm \frac{2a}{\pi}

B.  \fn_cm \frac{4a}{\pi}

C.  \fn_cm \frac{a}{\pi}

D.  \fn_cm \frac{a}{2\pi}

Check Answer
A
View Explanation
 

In processd

Q.10  Moment of inertia of a rectangular plate with sides a,b about an axis perpendicular to plate and passing through vertix is

A. \fn_cm \frac{1}{3}Ma^2

B.  \fn_cm \frac{1}{3}Mb^2

C. \fn_cm \frac{1}{3}M(a^2-b^2)

D.  \fn_cm \frac{1}{3}M(a^2+b^2)

Check Answer
D
View Explanation
 

In process

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